Is ${985959}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {985959}= &&{9}\cdot100000+ \\&&{8}\cdot10000+ \\&&{5}\cdot1000+ \\&&{9}\cdot100+ \\&&{5}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {985959}= &&{9}(99999+1)+ \\&&{8}(9999+1)+ \\&&{5}(999+1)+ \\&&{9}(99+1)+ \\&&{5}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {985959}= &&\gray{9\cdot99999}+ \\&&\gray{8\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {9}+{8}+{5}+{9}+{5}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${985959}$ is divisible by $3$ if ${ 9}+{8}+{5}+{9}+{5}+{9}$ is divisible by $3$ Add the digits of ${985959}$ $ {9}+{8}+{5}+{9}+{5}+{9} = {45} $ If ${45}$ is divisible by $3$ , then ${985959}$ must also be divisible by $3$ Add the digits of ${45}$ $ {4}+{5} = \color{#9D38BD}{9} $ If $\color{#9D38BD}{9}$ is divisible by $3$ , then ${45}$ must also be divisible by $3$ $\color{#9D38BD}{9}$ is divisible by $3$, therefore ${985959}$ must also be divisible by $3$.